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Sveta_85 [38]
2 years ago
7

If an air parcel is given a small push upward and it falls back to its original position, the atmosphere is said to be__________

.
Physics
1 answer:
nydimaria [60]2 years ago
6 0

If an air parcel is given a small push upward and it falls back to its original position, the atmosphere is said to be Stable.

A Stable atmospheric condition is a state where a body thrown like an parcel tends to return to its original position where it was present earlier even after being disturbed externally.

Stable conditions tends to oppose the forces that lead to the displacement of the object from original position.

When a parcel is given a small push upward and it falls back to its original position in air. This shows Stability in the atmosphere which bring back the sir parcel where it was present initially.

A body under the action of gravity in case of stable equilibrium conditions shows returning back of the body after sometime on reaching a certain height.

Learn more about Gravity here, brainly.com/question/4014727

#SPJ4

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Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
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Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

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now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

η_{th = 1 - T_L/0.5T_L

η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

Q_H = 120 kJ

Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

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Q_L = 60 kJ

now, the amount of heat rejection per kg of steam is;

q_L = Q_L/m

we substitute

q_L = 60/0.025

q_L = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

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T_L = 40 + ( 45 - 40 ) ( \frac{2400-2406.0}{2394.0-2406.0}\\} )

T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

T_L = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

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